Chapter 7

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Support Reactions. Referring to Fig. a, a

a

Method of Sections. It is required that . Referring to Fig. b,

Therefore,

Ans.

a Ans.

Also, . Referring to Fig. c,

Therefore,

Ans.

a Ans.

Method of Joints.

Joint A: Referring to Fig. d,

Ans.

Joint B: Referring to Fig. e,

Ans. Joint C: Ans. FCD = 30.0 kN (C) 40 - 14.14 sin 45° - FCD = 0 + ca F_y = 0; FBE = 20.0 kN (C) 14.14 sin 45° + 14.14 sin 45° - FBE = 0 + ca F_y = 0; FAF= 60.0 kN (C) 70 - 14.14 sin 45° - FAF= 0 + ca F_y = 0; FBC= 10.0 kN (T) 14.14 cos 45°(3) - FBC(3) = 0 +a M_D = 0; FDE = 10.0 kN (C) 14.14 cos 45°(3) - FDE(3) = 0 +a M_C = 0; FCE = 14.1 kN (C) FBD = 14.1 kN (T) F2 = 14.14 kN 40 - 20 - 2F2sin 45° = 0 + ca F_y = 0; FBD = F_CE = F₂ FAB= 10.0 kN (T) FAB(3) - 14.14 cos 45°(3) = 0 +a M_F = 0; FEF = 10.0 kN (C) FEF(3) - 14.14 cos 45°(3) = 0 +a M_A = 0; FAE = 14.1 kN (C) FBF = 14.1 kN (T) F1 = 14.14 kN 70 - 50 - 2F1sin 45° = 0 + ca F_y = 0; FBF = F_AE = F₁ Ax = 0 + : a Fx = 0; Ay = 70 kN 40(3) + 50(6) - Ay(6) = 0 +a M_C = 0; Cy = 40 kN Cy(6) - 40(3) - 20(6) = 0 +a M_A = 0;

7–1. Determine (approximately) the force in each member of the truss. Assume the diagonals can support either a tensile or a compressive force.

3 m 3 m 3 m 50 kN A D B C 40 kN 20 kN F E

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7–1. Contiuned

7–2. Solve Prob. 7–1 assuming that the diagonals cannot support a compressive force.

Support Reactions. Referring to Fig. a, a

a

Method of Sections. It is required that

Ans. FAE = F_CE = 0 Ax = 0 + : a Fx = 0; Ay = 70 kN 40(3) + 50(6) - Ay(6) = 0 +a M_C = 0; Cy = 40 kN Cy(6) - 40(3) - 20(6) = 0 +a M_A = 0; 3 m 3 m 3 m 50 kN A D B C 40 kN 20 kN F E

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2 0 2 Referring to Fig. b, Ans. a Ans. a Ans. Referring to Fig. c, Ans. a ; Ans. a Ans. Method of Joints.

Joint A: Referring to Fig. d,

Ans.

Joint B: Referring to Fig. e,

Ans.

Joint C: Referring to Fig. f,

Ans. FCD = 40.0 kN (C) 40 - FCD = 0 + ca F_y = 0; FBE = 40.0 kN (C) 28.28 sin45° + 28.28 sin45° - FBE = 0 + ca F_y = 0; FAF = 70.0 kN (C) 70 - FAF = 0 + ca F_y = 0; FBC = 0 -F_BC(3) = 0 +a M_D = 0; FDE = 20.0 kN (C) 28.28 cos 45°(3) - FDE(3) = 0 +a M_C = 0 FBD = 28.28 kN (T) = 28.3 kN (T) 40 - 20 - FBD sin 45° = 0 + ca F_y = 0; FAB= 0 FAB(3) = 0 +a M_F = 0 FEF = 20.0 kN (C) FEF(3) - 28.28 cos 45°(3) = 0 +a M_A = 0; FBF= 28.28 kN (T) = 28.3 kN (T) 70 - 50 - FBFsin 45° = 0 + ca F_y = 0;

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Assume is carried equally by and , so Ans. Ans. Joint A: Ans. Ans. Joint H: Ans. Ans. Ans. Joint G: Ans. Ans. FGB = 5.0 k (C) -10 + F_GB + 5.89 sin 45° + 1.18 sin 45° = 0 + ca F_y = 0; FGF = 7.5 k (C) 4.17 + 5.89 cos 45° - 1.18 cos 45° - FGF = 0 :+ a Fx = 0; FBF = 1.667 2 cos 45° = 1.18 k (T) FGC = 1.667 2 cos 45° = 1.18 k (C) VPanel = 1.667 k FHG = 4.17 k (C) -F_HG + 5.89 cos 45° = 0; :+ a Fx = 0; FAH = 14.16 k (C) -F_AH + 18.33 - 5.89 sin 45° = 0; + ca F_y = 0; FAB= 9.17 k (T) FAB- 5 - 5.89 cos 45° = 0; :+ a Fx = 0; FAG = 8.33 2 cos 45° = 5.89 k (C) FHB = 8.33 2 cos 45° = 5.89 k (T) FAG FHB VPanel VPanel = 8.33 k 20 ft 20 ft 20 ft 20 ft 10 k H A _D B C 10 k G 10 k F 10 k 5 k E

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2 0 4 Joint B: Ans. Ans. Ans. Joint D: Ans. Ans. Joint E: Ans. Joint C: Ans. FFC = 5.0 k (C) -F_FC + 8.25 sin 45° - 1.18 sin 45° = 0 + ca F_y = 0; FFE = 0.833 k (C) 5 + FFE - 8.25 cos 45° = 0 :+ a Fx = 0; FED = 15.83 k (C) 21.667 - 8.25 sin45° - FED = 0 + ca F_y = 0; FCD= 8.25 cos 45° = 5.83 k (T) :+ a Fx = 0; FDF = 11.567 2 cos 45° = 8.25 k (C) FEC = 11.667 2 cos 45° = 8.25 k (T) VPanel = 21.667 - 10 = 11.667 k FBC = 12.5 k (T) FBC+ 1.18 cos 45° - 9.17 - 5.89 cos 45° = 0 :+ a Fx = 0;

7–3. Continued Ans. Ans. Joint A: Ans. :+ a Fx = 0; F_AB= 5 k (T) FHB = 8.33 sin 45° = 11.785 = 11.8 k FAG = 0 VPanel = 8.33 k

*7–4. Solve Prob. 7–3 assuming that the diagonals cannot support a compressive force.

20 ft 20 ft 20 ft 20 ft 10 k H A _D B C 10 k G 10 k F 10 k 5 k E

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Ans. Joint H: Ans. Ans. Ans. Joint B: Ans. Ans. Joint G: Ans. Ans. Ans. Joint D: Ans. Ans. Joint E: Ans. Joint F: Ans. FFC = 11.7 k (C) + ca F_y = 0; F_FC - 10 - 2.36 sin 45° = 0 FEF = 6.67 k (C) FEF + 5 - 16.5 cos 45° = 0 :+ a Fx = 0; + ca F_y = 0; F_ED = 21.7 k (C) FCD= 0 :+ a Fx = 0; FEC = 11.667 sin 45° = 16.5 k (T) FDF = 0 VPanel = 11.667 k FGF = 8.33 k (C) :+ _{a F}_x = 0; FGB = 10 k (C) - F_GB + 11.785 sin45° + 2.36 sin 45° = 0 + ca F_y = 0; FBC = 11.7 k (T) FBC+ 2.36 cos 45° - 11.785 cos 45° - 5 = 0 :+ a Fx = 0; FBF = 1.667 sin 45° = 2.36 k (T) FGC = 0 VPanel = 1.667 k FHG = 8.33 k (C) 11.785 cos 45° - F_HG = 0 :+ a Fx = 0; + ca F_y = 0; F_AN = 18.3 k (C) 7–4. Continued

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2 0 6 Support Reactions. Referring to, Fig. a

a

Method of Sections. It is required that . Referring to Fig. b,

Therefore,

Ans.

a Ans.

It is required that . Referring to Fig. c,

+ ca F_y = 0; 21.5 - 7 - 14 - 2F₂a3 5b = 0 F2 = 0.4167 k FCG = F_BF = F₂ +a M_A = 0; F_GH(6) - 12.08a 4 5b(6) = 0 FGH = 9.667 k (C) = 9.67 k (C) +a M_H = 0; F_AB(6) + 2(6) - 12.08a4 5b(6) = 0 FAB = 7.667 k (T) = 7.67 k (T) FBH = 12.1 k (T) F_AG = 12.1 k (C) + ca F_y = 0; 21.5 - 7 - 2F₁a 3 5b = 0 F1= 12.08 k FBH= F_AG = F₁ +a M_D = 0; 14(8) + 14(16) + 7(24) + 2(6) - A_y(24) = 0 A_y = 21.5 k +a M_A = 0; D_y(24) + 2(6) - 7(24) - 14(16) - 14(8) = 0 D_y = 20.5 k :+ a Fx = 0; A_x - 2 = 0 A_x = 2 k

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Therefore, Ans. a Ans. a Ans. It is required that . Referring to Fig. d

Therefore,

Ans.

a Ans.

Method of Joints.

Joint A: Referring to Fig. e,

Ans.

Joint B: Referring to Fig. f,

Ans. + ca F_y = 0; 12.08a3 5b - 0.4167a 3 5b - FBG = 0 FBG = 7.00 k (C) + ca F_y = 0; 21.5 - 12.08a 3 5b - FAH = 0 FAH = 14.25 k (C) +a M_E = 0; 11.25(0.8)(6) - F_CD(6) = 0 F_CD= 9.00 k (T) +a M_D = 0; 2(6) + 11.25a4 5b(6) - FEF(6) = 0 FEF = 11.0 k (C) FCE = 11.25 k (T) F_DF= 11.25 k (C) + ca F_y = 0; 20.5 - 7 - 2F₃a 3 5b = 0 F3= 11.25 k FCE = F_DF= F₃ FBC = 17.67 k (T) = 17.7 k (T) +a M_G = 0; F_BC(6) + 7(8) + 2(6) - 21.5(8) - 0.4167a4 5b(6) = 0 FFG = 19.67 k (C) = 19.7 k (C) +a M_B = 0; F_FG(6) - 0.4167a4 5b(6) + 7(8) - 21.5(8) = 0 FCG = 0.417 k (T) F_BF = 0.417 k (C) 7–5. Continued

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2 0 8

7–5. Continued

Support Reactions. Referring to Fig. a,

a

Method of Sections. It is required that

Ans. Referring to Fig. b, Ans. a Ans. a+a M_H = 0; 2(6) - F_AB(6) = 0 F_AB = 2.00 k (C) Ans. +a M_A = 0; F_GH(6) - 24.17a 4 5b(6) = 0 FGH = 19.33 k (C) = 19.3 k (C) + ca F_y = 0; 21.5 - 7 - F_BHa3 5b = 0 FBH = 24.17 k (T) = 24.2 k (T) FAG = F_BF = F_DF = 0 +a M_D = 0; 14(8) + 14(16) + 7(24) + 2(6) - A_y(24) = 0 A_y = 21.5 k +a M_A = 0; D_y(24) + 2(6) - 7(24) - 14(16) - 14(8) = 0 D_y = 20.5 k :+ a Fx = 0; A_x - 2 = 0 A_x = 2 k

7–6. Solve Prob. 7–5 assuming that the diagonals cannot support a compressive force.

Joint D: Referring to Fig. h,

Ans. FDE = 13.75 k

+ ca F_y = 0; 20.5 - 11.25a3

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Referring to Fig. c Ans. a Ans. a Ans. Referring to Fig. d, Ans. a Ans. a Ans. Method of Joints.

Joint A: Referring to Fig. e,

Ans. Joint B: Referring to Fig. f,

Ans. Joint C: Referring to Fig. g,

Ans. Joint D: Referring to Fig. h,

Ans. + ca F_y = 0; 20.5 - F_DE = 0 F_DE = 20.5 k (C) + ca F_y = 0; 0.8333a 3 5b + 22.5a 3 5b - FCF = 0 FCF = 14.0 k (C) + ca F_y = 0; 24.17a 3 5b - FBG = 0 FBG = 14.5 k (C) + ca F_y = 0; 21.5 - F_AH = 0 F_AH = 21.5 k (C) +a M_E = 0; -F_CD(6) = 0 F_CD= 0 +a M_D = 0; 2(6) + 22.5a4 5b(6) - FEF(6) = 0 FEF = 20.0 kN (C) + ca F_y = 0; 20.5 - 7 - F_CEa3 5b = 0 FCE = 22.5 k (T) +a M_G = 0; F_BC(6) + 7(8) + 2(6) - 21.5(8) = 0 F_BC = 17.33 k (T) = 17.3 k (T) FFG = 20.0 k (C) +a M_B = 0; F_FG(6) + 7(8) - 21.5(8) - 0.8333a4 5b(6) = 0 + ca F_y = 0; 21.5 - 7 - 14 - F_CGa3 5b = 0 FCG = 0.8333 k (T) = 0.833 k (T) 7–6. Continued

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2 1 0

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Assume Ans. Ans. a Ans. Ans. Joint C: Ans. Assume Ans. Ans. a Ans. Ans. Joint B: Ans. Joint A: Ans. FAF = 6.00 kN (T) + ca F_y = 0; F_AF - 10.0a 1.5 2.5b = 0 FBE = 4.00 kN (T) + ca F_y = 0; F_BE + 10.0a 1.5 2.5b - 3.333a 1.5 2.5b - 8 = 0 :+ a Fx = 0; F_AB= 13.3 kN (C) FFE = 13.3 kN (T) +a M_B = 0; F_FE(1.5) - 10.0a 2 2.5b(1.5) - 4(2) = 0 FAE = 10.0 kN (C) FFB = 10.0 kN (T) + ca F_y = 0; 2F_FBa1.5 2.5b - 8 - 4 = 0 FFB = F_AE FCD = 2.00 kN (T) + ca F_y = 0; F_CD + 3.333a1.5 2.5b - 4 = 0 :+ _{a F}_x = 0; F_BC= 2.67 kN (C) FED = 2.67 kN (T) +a M_C = 0; F_ED(1.5) - a 2 2.5b(3.333)(1.5) = 0 FBD = 3.333 kN = 3.33 kN (C) FEC = 3.333 kN = 3.33 kN (T) + ca F_y = 0; 2F_ECa1.5 2.5b - 4 = 0 FBD= F_EC

7–7. Determine (approximately) the force in each member of the truss. Assume the diagonals can support either a tensile or compressive force.

8 kN 4 kN

F E D

1.5 m

A B C

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2 1 2 Assume Ans. Ans. a Ans. Ans. Joint D: From Inspection: Ans. Assume Ans. Ans. a Ans. Ans. Joint B: Ans. Joint A: Ans. + ca F_y = 0; F_AF = 0 FBE = 4.00 kN (T) + ca F_y = 0; - F_BE - 8 + 20.0a1.5 2.5b = 0 FAB = 21.3 kN (C) :+ a Fx = 0; F_AB - 5.333 - 20.0a 2 2.5b = 0 FFE = 5.333 kN = 5.33 kN (T) +a M_B = 0; F_FE(1.5) - 4(2) = 0 FFB = 20.0 kN (T) + ca F_y = 0; F_FBa 1.5 2.5b - 8 - 4 = 0 FAE = 0 FCD = 0 FBC = 5.33 kN (C) :+ _{a F}_x = 0; F_BC - 6.667a 2 2.5b = 0 +a M_C = 0; F_ED = 0 FEC = 6.667 kN = 6.67 kN (T) + ca F_y = 0; F_ECa1.5 2.5b - 4 = 0 FBD = 0

*7–8. Solve Prob. 7–7 assuming that the diagonals cannot support a compressive force.

8 kN 4 kN

F E D

1.5 m

A B C

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Method of Sections. It is required that . Referring to Fig. a, Therefore, Ans. a Ans. a Ans. It is required that . Referring to Fig. b,

Therefore, Ans. a Ans. a Ans. Method of Joints.

Joint E: Referring to Fig. c,

Ans.

Joint F: Referring to Fig. d,

Ans. FDF = 0.250 k (C) :+ a Fx = 0; 2.475 sin 45° - 2.121 cos 45° - F_DF = 0 + ca F_y = 0; 2.121 sin 45° - F_DE = 0 F_DE = 1.50 k (T) :+ a Fx = 0; F_EFcos45° - 1.5 = 0 F_EF = 2.121 k (C) = 2.12 k (C) FAG = 7.75 k (C) +a M_C = 0; 1.5(30) + 2(15) + 3.889 cos 45° (15) - F_AG(15) = 0 FBC = 7.75 k (T) +a M_G = 0; 1.5(30) + 2(15) + 3.889 cos 45° (15) - F_BC(15) = 0 FBG = 3.89 k (T) F_AC = 3.89 k (C) :+ a Fx = 0; 2F₂sin 45° - 2 - 2 - 1.5 = 0 F₂ = 3.889 k FBG = F_AC = F₂ FCD = 3.25 k (T) +a M_F = 0; 1.5(15) + 2.475 cos 45° (15) - F_CD(15) = 0 FFG = 3.25 k (C) +a M_D = 0; 1.5(15) + 2.475 cos 45° (15) - F_FG(15) = 0 FCF = 2.48 k (T) F_DG = 2.48 k (C) :+ _{a F}_x = 0; 2F₁ sin 45° - 2 - 1.5 = 0 F₁ = 2.475 k FCF= F_DG = F₁ 7–9. Determine (approximately) the force in each member of the truss. Assume the diagonals can support both tensile and compressive forces.

15 ft 15 ft 2 k 2 k 1.5 k 15 ft 15 ft E F A B C G D

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2 1 4 Joint G: Referring to Fig. e,

Ans. Joint A: Referring to Fig. f,

Ans. FAB = 2.75 k :+ a Fx = 0; F_AB - 3.889 cos 45° = 0 FCG = 1.00 k (C) :+ a Fx = 0; 3.889 sin 45° - 2.475 cos 45° - F_CG = 0 7–9. Continued

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Method of Sections. It is required that Ans. Referring to Fig. a, Ans. a Ans. a Ans. Referring to Fig. b, Ans. a Ans. a Ans. Method of Joints.

Ans.

Ans. Joint F: Referring to Fig. d,

Ans. Joint G: Referring to Fig. e,

Ans. Joint A: Referring to Fig. f,

Ans. :+ a Fx = 0; F_AB= 0 :+ a Fx = 0; 7.778 sin 45° - F_CG = 0 F_CG = 5.50 k (C) :+ a Fx = 0; 4.950 sin 45° - 2.121 cos 45° - F_DF= 0 F_DF= 2.00 k (C) + ca F_y = 0; 2.121 sin 45° - F_DE = 0 F_DE = 1.50 k (T) :+ _{a F}_x = 0; F_EFcos 45° - 1.5 = 0 F_EF = 2.121 k (C) = 2.12 k (C) FAG = 10.5 k (C) +a M_C = 0; 1.5(30) + 2(15) + 7.778 cos 45° - F_AG(15) = 0 +a M_G = 0; 1.5(30) + 2(15) - F_BC(15) = 0 F_BC= 5.00 k (T) :+ a Fx = 0; F_BGsin 45° - 2 - 2 - 1.5 = 0 F_BG = 7.778 k (T) = 7.78 k (T) FFG = 5.00 k (C) +a M_D = 0; 1.5(15) + 4.950 cos 45°(15) - F_FG(15) = 0 +a M_F = 0; 1.5(15) - F_CD(15) = 0 F_CD = 1.50 k (T) :+ a Fx = 0; F_CFsin 45° - 1.5 - 2 = 0 F_CF = 4.950 k (T) = 4.95 k (T) FDG = F_AC = 0

7–10. Determine (approximately) the force in each member of the truss. Assume the diagonals DG and AC cannot support a compressive force.

15 ft 15 ft 2 k 2 k 1.5 k 15 ft 15 ft E F A B C G D

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2 1 6 7–10. Continued

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Method of Sections. It is required that . Referring to Fig. a, Therefore, Ans. a Ans. a Ans.

It is required that Referring to Fig. b,

Therefore, Ans. a Ans. a Ans. Method of Joints.

Joint D: Referring to Fig. c,

Ans. Joint C: Referring to Fig. d,

Ans. Joint B: Referring to Fig. e,

Ans. :+ a Fx = 0; 15.0a3 5b - FAB= 9.00 kN (T) :+ a Fx = 0; F_CF+ 6.667a 3 5b - 15.0a 3 5b = 0 FCF = 5.00 kN (C) :+ a Fx = 0; F_DE - 6.667a 3 5b = 0 FDE = 4.00 kN (C) FAF = 22.67 kN (T) = 22.7 kN (T) +a M_C = 0; F_AF(1.5) - 15.0a 4 5b(1.5) - 8(2) = 0 FBC = 22.67 kN (C) = 22.7 kN (C) +a M_F = 0; F_BC(1.5) - 15.0a 4 5b(1.5) - 8(2) = 0 FBF = 15.0 kN (C) F_AC = 15.0 kN (T) :+ _{a F}_x = 0; 8 + 10 - 2F₂a3 5b = 0 F2 = 15.0 kN FBF = F_AC = F₂ +a M_D = 0; F_EF(1.5) - 6.667a4 5b(1.5) = 0 FEF = 5.333 kN (T) = 5.33 kN (T) +a M_E = 0; F_CD(1.5) - 6.667a4 5b(1.5) = 0 FCD = 5.333 kN (C) = 5.33 kN (C) FCE = 6.67 kN (C) F_DF = 6.67 kN (T) :+ a Fx = 0; 8 - 2F₁a3 5b = 0 F1= 6.667 kN FCE = F_DF= F₁ 7–11. Determine (approximately) the force in each member of the truss. Assume the diagonals can support

either a tensile or compressive force. 8 kN

1.5 m E F A B C D 10 kN 2 m 2 m

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2 1 8 7–11. Continued

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Method of Sections. It is required that Ans. Referring to Fig. a, Ans. a Ans. a Ans. Referring to Fig. b, Ans. a Ans. a Ans. Method of Joints.

Ans. Joint C: Referring to Fig. d,

Ans. Joint B: Referring to Fig. e,

Ans. :+ a Fx = 0; F_AB= 0 :+ a Fx = 0; F_CF - 30.0a 3 5b = 0 FCF = 18.0 kN (C) :+ a Fx = 0; 8 - F_DE = 0 F_DE = 8.00 kN (C) FBC = 34.67 kN (C) = 34.7 kN (C) +a M_F = 0; F_BC(1.5) - 30.0a 4 5b(1.5) - 8(2) = 0 +a M_C = 0; F_AF(1.5) - 8(2) = 0 F_AF= 10.67 kN (T) :+ _{a F}_x = 0; 8 + 10 - F_ACa3 5b = 0 FAC = 30.0 kN (T) +a M_D = 0; F_EF(1.5) = 0 F_EF = 0 +a M_E = 0; F_CD(1.5) - 13.33a4 5b(1.5) = 0 FCD = 10.67 kN (C) = 10.7 kN (C) :+ a Fx = 0; 8 - F_DFa3 5b = 0 FDF = 13.33 kN (T) = 13.3 kN (T) FCE = F_BF = 0

*7–12. Determine (approximately) the force in each member of the truss. Assume the diagonals cannot support

a compressive force. 8 kN 1.5 m E F A B C D 10 kN 2 m 2 m

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2 2 0 7–12. Continued

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The frame can be simplified to that shown in Fig. a, referring to Fig. b, a Ans. Referring to Fig. c, a Ans. MB = 3.78 kN

#

m +a M_B = 0; M_B - 9.6(0.8) - 3(1.4)(0.1) + 7.2(0.6) = 0 +a M_A = 0; M_A - 7.2(0.6) - 3(0.6)(0.3) = 0 M_A = 4.86 kN

#

m 7–13. Determine (approximately) the internal moments

at joints A and B of the frame.

6 m A B C D E F G H 8 m 6 m 6 m 3 kN/m

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2 2 2 a Ans. a Ans. MD = 7.20 k

#

ft +a M_D = 0; - M_D + 0.8(1) + 3.2(2) = 0 MF = 4.05 k

#

ft +a M_F = 0; M_F - 0.6(0.75) - 2.4(1.5) = 0 7–14. Determine (approximately) the internal moments at joints F and D of the frame.

400 lb/ft F E A B C D 15 ft 20 ft

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The frame can be simplified to that shown in Fig. a, Referring to Fig. b, a

Ans. MA= 40.32 kN

#

m = 40.3 kN

#

m

+a M_A = 0; M_A - 5(0.8)(0.4) - 16(0.8) - 9(0.8)(0.4) - 28.8(0.8) = 0 7–15. Determine (approximately) the internal moment at

A caused by the vertical loading.

8 m A C E B D F 5 kN/m 9 kN/m

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2 2 4 The frame can be simplified to that shown in Fig. a. The reactions of the 3 kN/m and 5 kN/m uniform distributed loads are shown in Fig. b and c respectively. Referring to Fig. d,

a Ans. Referring to Fig. e, a Ans. MB = 14.4 kN

#

m 9.60(0.8) - 9.60(0.8) + 5(0.8)(0.4) + 16(0.8) - MB = 0 +a M_B = 0; MA = 23.04 kN

#

m = 23.0 kN

#

m MA - 3(0.8)(0.4) - 9.6(0.8) - 5(0.8)(0.4) - 16(0.8) = 0 +a M_A = 0;

*7–16. Determine (approximately) the internal moments at A and B caused by the vertical loading.

A B D E F K L G H I J C 5 kN/m 5 kN/m 3 kN/m 8 m 8 m 8 m

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Joint I: a Ans. Joint L: a Ans. Joint H: a Ans. MH = 27.0 k

#

ft MH - 3.0(1) - 12.0(2) = 0 +a M_H = 0; ML = 20.25 k

#

ft ML - 6.0(3) - 1.5(1.5) = 0 +a M_L = 0; MI = 9.00 k

#

ft MI - 1.0(1) - 4.0(2) = 0 +a M_I = 0;

7–17. Determine (approximately) the internal moments at joints I and L. Also, what is the internal moment at joint

H caused by member HG? 20 ft 40 ft 30 ft A H I B G F K L E D J C 0.5 k/ft 1.5 k/ft 1.5 k/ft

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2 2 6 Ans. Ans. Ans. MC = 7.2 k

#

ft MB = 9 k

#

ft MA = 16.2 k

#

ft Cy = 4 k By = 16 k Ay = 12 k Cx = 0 Bx = 0 Ax = 0

7–18. Determine (approximately) the support actions at

A, B, and C of the frame.

15 ft 20 ft A D F C H G E B 400 lb/ft 1200 lb/ft

(28)

7–19. Determine (approximately) the support reactions at A and B of the portal frame. Assume the supports are (a) pinned, and (b) fixed.

For printed base, referring to Fig. a and b,

a (1)

a (2)

Solving Eqs. (1) and (2) yield

Referring to Fig. a, Ans. Ans. Referring to Fig. b, Ans. Ans. For the fixed base, referring to Fig. c and d,

a (1)

a (2)

Solving Eqs (1) and (2) yields,

Referring to Fig. c, Referring to Fig. d, Gy = 9.00 kN Gy - 9.00 = 0 + ca F_y = 0; Gx = 6.00 kN 6.00 - Gx = 0 :+ a Fx = 0; Ey = 9.00 kN 9.00 - Ey = 0 + ca F_y = 0; Ex = 6.00 kN 12 - 6.00 - Ex = 0 :+ a Fx = 0; Fx = 6.00 kN Fy = 9.00 kN Fy(2) - Fx(3) = 0 +a M_G = 0; Fx(3) + Fy(2) - 12(3) = 0 +a M_E = 0; By = 18.0 kN By - 18.0 = 0 + ca F_y = 0; Bx = 6.00 kN 6.00 - Bx = 0 :+ _{a F}_x = 0; Ay = 18.0 kN 18.0 - Ay = 0 + ca F_y = 0; Ax = 6.00 kN 12 - 6.00 - Ax = 0 :+ _{a F}_x = 0; Ex = 6.00 kN Ey = 18.0 kN Ey(6) - Ex(6) = 0 +a M_B = 0; Ex(6) + Ey(2) - 12(6) = 0 +a M_A = 0; 6 m 4 m 12 kN A D C B

(29)

2 2 8 Referring to Fig. e, Ans. Ans. a Ans. Referring to Fig. f, Ans. Ans. a+a M_B = 0; M_B - 6.00(3) = 0 MB = 18.0 kN

#

m Ans. By = 9.00 kN By - 9.00 = 0 + ca F_y = 0; Bx = 6.00 kN 6.00 - Bx = 0 :+ a Fx = 0; MA = 18.0 kN

#

m MA - 6.00(3) = 0 +a M_A = 0; Ay = 9.00 kN 9.00 - Ay = 0 + ca F_y = 0; Ax = 6.00 kN 6.00 - Ax = 0 :+ a Fx = 0;

(30)

a Ans. Ans. Member BC: Ans. Members AB and CD: Ans. VA= V_B = V_C = V_D = P 2 VB= V_C = 2Ph 3b MB = M_C= P 2a 2h 3 b = Ph 3 MA = M_D = P 2a h 3b = Ph 6 Ey = 2Ph 3b Ey = 2Ph 3b = 0 + ca F_y = 0; Gy = Pa2h 3bb Gy(b) - Pa 2h 3 b = 0 +a M_B = 0;

*7–20. Determine (approximately) the internal moment and shear at the ends of each member of the portal frame. Assume the supports at A and D are partially fixed, such that an inflection point is located at h/3 from the bottom of each column. P B A C D b h

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2 3 0

Inflection points are at A and B. From FBD (1): a From FBD (2): a Ans. Ex = 1500 lb -250 + 1767.8(sin 45°) + 500 - E_x = 0 :+ a Fx = 0; FCF = 1.77 k(T) 250(7.5) - FCF(sin 45°)(1.5) = 0; +a M_E = 0; Ay = 375 lb Ay(10) - 500(7.5) = 0; +a M_B = 0;

7–21. Draw (approximately) the moment diagram for member ACE of the portal constructed with a rigid member

EG and knee braces CF and DH. Assume that all points of

connection are pins. Also determine the force in the knee brace CF.

*7–22. Solve Prob. 7–21 if the supports at A and B are fixed instead of pinned.

Inflection points are as mid-points of columns a a Ans. Ex = 1.00 k 500 + 1060.66(sin 45°) - 250 - Ex = 0; :+ a Fx = 0; FCE = 1.06 k(T) 250(4.5) - FCE(sin 45°)(1.5) = 0; +a M_E = 0; Iy = 175 lb -I_y+ 175 = 0; + ca F_y = 0; Jy = 175 lb Jy(10) - 500(3.5) = 0; +a M_I= 0; 7 ft 500 lb B A E G H F D C 6 ft 1.5 ft 1.5 ft 1.5 ft 7 ft 500 lb B A E G H F D C 6 ft 1.5 ft 1.5 ft 1.5 ft

(32)

7–23. Determine (approximately) the force in each truss member of the portal frame. Also find the reactions at the fixed column supports A and B. Assume all members of the truss to be pin connected at their ends.

Assume that the horizontal reactive force component at fixed supports A and B are equal. Thus

Ans. Also, the points of inflection H and I are at 6 ft above A and B respectively. Referring to Fig. a, a Referring to Fig. b, Ans. a Ans. Referring to Fig. c, Ans. a Ans.

Using the method of sections, Fig. d,

Ans.

a Ans.

Using the method of Joints, Fig. e,

Ans. FDF = 3.125 k (T) FDFa 3 b - 3.125a3b = 0 + ca F_y = 0; FFG = 1.00 k (C) FFG(6) - 2(6) + 1.5(6) + 1.875(8) = 0 +a M_D= 0; FCD = 2.00 k (C) FCD(6) + 1(6) - 1.50(12) = 0 +a M_G = 0; FDG = 3.125 k (C) FDGa 3 5b - 1.875 = 0 + ca F_y = 0; MB = 9.00 k

#

ft MB - 1.50(6) = 0 +a M_B= 0; By = 1.875 k By - 1.875 = 0 + ca F_y = 0; Bx = 1.50 k 1.50 - Bx = 0 :+ a Fx = 0; MA= 9.00 k

#

ft MA - 1.50(6) = 0 +a M_A= 0; Ay = 1.875 k 1.875 - Ay = 0 + ca F_y = 0; Hx = 1.50 k Hx - 1.50 = 0 :+ a Fx = 0; Iy = 1.875 k Iy - 1.875 = 0 + ca F_y = 0; Hy = 1.875 k Hy(16) - 1(6) - 2(12) = 0 +a M_I= 0; Ax = B_x = 2 + 1 2 = 1.50 k 8 ft 2 k 8 ft 6 ft 12 ft A B C D F E G 1 k

(33)

2 3 2

7–23. Continued

*7–24. Solve Prob. 7–23 if the supports at A and B are pinned instead of fixed.

Assume that the horizontal reactive force component at pinal supports

A and B are equal. Thus,

Ans. Referring to Fig. a, a Ans. Ans. By = 3.00 k By - 3.00 = 0 + ca F_y = 0; Ay = 3.00 k Ay(16) - 1(12) - 2(18) = 0 +a M_B = 0; Ax = B_x = H2 2 = 1.50 k 8 ft 2 k 8 ft 6 ft 12 ft A B C D F E G 1 k

(34)

Using the method of sections and referring to Fig. b,

Ans.

a Ans.

Using the method of joints, Fig. c,

Ans. Ans. FDE = 4.50 k (C) 5.00a4 5b + 5.00a 4 5b - 3.50 - FDE = 0 :+ a Fx = 0; FDF = 5.00 k (T) FDFa 3 5b - 5.00a 3 5b = 0 + ca F_y = 0; FCD = 3.50 k (T) FCD(6) + 1(6) - 1.50(18) = 0 +a M_G = 0; FGF = 1.00 k (C) FGF(6) - 2(6) - 1.5(12) + 3(8) = 0 +a M_D = 0; FDG = 5.00 k (C) FDGa 3 5b - 3.00 = 0 + ca F_y = 0; 7–24. Continued

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2 3 4

Assume that the horizontal force components at pin supports A and B are equal. Thus,

Referring to Fig. a, a

Using the method of sections, Fig. b,

Ans. a Ans. a+a M_E = 0; 8(1.5) + 16.5(2) - 6(6.5) - F_CG(1.5) = 0 FCG = 4.00 kN (C) Ans. FEF = 24.0 kN (C) FEF(1.5) - 4(1.5) - 6.00(5) = 0 +a M_G = 0; FEG = 27.5 kN (T) FEGa 3 5b - 16.5 = 0 + ca F_y = 0; Ay = 16.5 kN 16.5 - Ay = 0 + ca F_y = 0; By = 16.5 kN By(4) - 8(5) - 4(6.5) = 0 +a M_A= 0; Ax = B_x = 4 + 8 2 = 6.00 kN

7–25. Draw (approximately) the moment diagram for column AGF of the portal. Assume all truss members and the columns to be pin connected at their ends. Also determine the force in all the truss members.

2 m 2 m 1.5 m 5 m A B D G F C E 4 kN 8 kN

(36)

7–26. Draw (approximately) the moment diagram for column AGF of the portal. Assume all the members of the truss to be pin connected at their ends. The columns are fixed at A and B. Also determine the force in all the truss members.

7–25. Continued

Assume that the horizontal force components at fixed supports A and B are equal. Thus,

Also, the points of inflection H and I are 2.5 m above A and B, respectively. Referring to Fig. a, a I 9.00 kN I 9.00 = 0 a F 0; Hy = 9.00 kN Hy(4) - 8(2.5) - 4(4) = 0 +a M_I= 0; Ax = B_x = 4 + 8 2 = 6.00 kN Using the method of joints, Fig. c,

Ans. Ans. FDE = 20.0 kN (T) 24 - 27.5a4 5b - 27.5a 4 5b + FDE = 0 :+ a Fx = 0; FCE = 27.5 kN (C) FCEa 3 5b - 27.5a 3 5b = 0 + ca F_y = 0; 2 m 2 m 1.5 m 5 m A B D G F C E 4 kN 8 kN

(37)

2 3 6

Referring to Fig. b,

a

Ans. a Ans. a+a M_G = 0; F_EF(1.5) - 4(1.5) - 6(2.5) = 0 FEF = 14.0 kN (C) Ans. FCG = 4.00 kN (C) 8(1.5) + 9.00(2) - 6.00(4) - FCG(1.5) = 0 +a M_E = 0; FEG= 15.0 kN(T) FEGa 3 5b - 9.00 = 0 + ca F_y = 0; MA = 15.0 kN

#

m MA - 6.00(2.5) = 0 +a M_A = 0; Ay = 9.00 kN 9.00 - Ay = 0 + ca F_y = 0; Hx = 6.00 kN Hx - 6.00 = 0 :+ _{a F}_x = 0; 7–26. Continued

(38)

7–27. Determine (approximately) the force in each truss member of the portal frame. Also find the reactions at the fixed column supports A and B. Assume all members of the truss to be pin connected at their ends.

7–26. Continued

Ans.

Also, the points of inflection J and K are 3 m above A and B respectively. Referring to Fig. a,

J J

Ax = B_x = 12 + 8

2 = 10.0 kN Using the method of joints, Fig. e,

Ans. Ans. FDE = 10.0 kN (T) FDE + 14.0 - 15.0a4 5b - 15.0a 4 5b = 0 :+ a Fx = 0; FCE = 15.0 kN (C) FCEa 3 5b - 15.0a 3 5b = 0 + ca F_y = 0; 6 m 6 m 1.5 m 1.5 m 3 m 3 m 8 kN 8 kN A A F F II G G B B 2 m 2 m 1.5 m 1.5 m H H CC D D E E 3 m 3 m 3 m3 m 12 kN 12 kN

(39)

2 3 8

Referring to Fig. b, Ans. a Ans. Referring to Fig. c, Ans. a Ans.

Using the metod of sections, Fig. d,

Ans.

a Ans.

Using the method of joints, Fig. e (Joint H),

Ans.

Ans. Referring Fig. f (Joint E),

Ans.

Ans. Referring to Fig. g (Joint I),

Ans. Ans. :+ a Fx = 0; 17.5a3 5b + 17.5a 3 5b + 4.00 - FCI = 0 FCI = 25.0 kN (C) + ca F_y = 0; F_DIa4 5b - 17.5a 4 5b = 0 FDI = 17.5 kN (T) :+ a Fx = 0 F_DE + 16.5 - 17.5a 3 5b - 17.5a 3 5b = 0 FDE = 4.50 kN (T) + ca F_y = 0; F_EIa 4 5b - 17.5a 4 5b = 0 FEI = 17.5 kN (C) :+ a Fx = 0; 17.5a 3 5b + 17.5a 3 5b - 17.0 - FHI = 0 FHI = 4.00 kN (C) + ca F_y = 0; F_EHa 4 5b - 17.5a 4 5b = 0 FEH = 17.5 kN (T) +a M_F = 0; F_GH(2) + 8(2) - 10.0(5) = 0 F_GH = 17.0 kN (T) +a M_H = 0; F_EF(2) + 14.0(1.5) - 12(2) - 10.0(3) = 0 F_EF = 16.5 kN (C) + ca F_y = 0; F_FHa4 5b - 14.0 = 0 FFH= 17.5 kN (C) +a M_B = 0; M_B - 10.0(3) = 0 M_B = 30.0 kN

#

m + ca F_y = 0; B_y - 14.0 = 0 B_y = 14.0 kN :+ _{a F}_x = 0; B_x - 10.0 = 0 B_x = 10.0 kN +a M_A = 0; M_A - 10.0(3) = 0 M_A = 30.0 kN

#

m + ca F_y = 0; 14.0 - A_y = 0 A_y = 14.0 kN :+ a Fx = 0; J_x - 10.0 = 0 J_x = 10.0 kN 7–27. Continued

(40)

(41)

2 4 0

Assume that the horizontal force components at pin supports A and B are equal. Thus,

Ans. Referring to Fig. a,

a Ans.

Ans.

Ans. a

Ans.

a Ans.

Using method of joints, Fig. c (Joint H),

Ans.

Referring to Fig. d (Joint E),

Ans.

Referring to Fig. e (Joint I),

Ans. Ans. FCI = 40.0 kN (C) :+ _{a F}_x = 0; 30.0a3 5b + 30.0a 3 5b + 4.00 - FCI = 0 + ca F_y = 0; F_DIa4 5b - 30.0a 4 5b = 0 FDI = 30.0 kN (T) :+ _{a F}_x = 0; F_DE + 24.0 - 30.0a3 5b - 30.0a 3 5b = 0 FDE = 12.0 kN (T) + ca F_y = 0; F_EIa4 5b - 30.0a 4 5b = 0 FEI = 30.0 kN (C) :+ a Fx = 0; 30.0a3 5b + 30.0a 3 5b - 32.0 - FHI = 0 FHI = 4.00 kN (C) + ca F_y = 0; F_EHa 4 5b - 30.0a 4 5b = 0 FEH = 30.0 kN (T) +a M_F = 0; F_GH(2) + 8(2) - 10.0(8) = 0 F_GH = 32.0 kN (T) FEF = 24.0 kN (C) +a M_H = 0; F_EF(2) + 24.0(1.5) - 12(2) - 10.0(6) = 0 + ca F_y = 0; F_FHa4 5b - 24.0 = 0 FFH= 30.0 kN (C) + ca F_y = 0; B_y - 24.0 = 0 B_y = 24.0 kN +a M_B = 0; A_y(6) - 8(6) - 12(8) = 0 A_y = 24.0 kN Ax = B_x = 12 + 8 2 = 10.0 kN

*7–28. Solve Prob. 7–27 if the supports at A and B are pinned instead of fixed.

6 m 6 m 1.5 m 1.5 m 3 m 3 m 8 kN 8 kN A A F F II G G B B 2 m 2 m 1.5 m 1.5 m H H CC D D E E 3 m 3 m 3 m3 m 12 kN 12 kN

(42)

(43)

2 4 2

Ans. Also, the points of inflection N and O are at 6 ft above A and B respectively. Referring to Fig. a, a a Referring to Fig. b, Ans. a Ans. Referring to Fig. c, Ans. a Ans.

Ans. a Ans. a Ans. FJK = 0.500 k (C) +a M_G = 0; - F_JK(6) + 4(6) + 1.125(8) - 2.00(15) = 0 +a M_K = 0; F_GF(6) + 1.125(16) - 2(9) = 0 F_GF = 0 + ca F_y = 0; F_GKa3 5b - 1.125 = 0 FGK = 1.875 k (C) +a M_B = 0; M_B - 2.00(6) = 0 M_B = 12.0 k

#

ft + ca F_y = 0; B_y - 1.125 = 0 B_y = 1.125 k :+ a Fx = 0; B_x - 2.00 = 0 B_x = 2.00 k +a M_A = 0; M_A - 2.00(6) = 0 M_A = 12.0 k

#

ft + ca F_y = 0; 1.125 - A_y = 0 A_y = 1.125 k :+ a Fx = 0; N_x - 2.00 = 0 N_x = 2.00 k +a M_N = 0; O_y(32) - 4(9) = 0 O_y = 1.125 k +a M_B = 0; N_y(32) - 4(9) = 0 N_y = 1.125 k Ax = B_x = 4 2 = 2.00 k

7–29. Determine (approximately) the force in members

GF, GK, and JK of the portal frame. Also find the reactions

at the fixed column supports A and B. Assume all members of the truss to be connected at their ends.

D C E H I K F L J G 8 ft 8 ft 8 ft 8 ft 12 ft 3 ft 6 ft A 4 k B

(44)

Assume that the horizontal force components at pin supports A and B are equal. Thus, Ans. Referring to Fig. a, a Ans. Ans.

Ans. a Ans. a Ans. +a M_G = 0; 4(6) + 1.875(8) - 2.00(21) + F_JK(6) = 0 F_JK= 0.500 k (T) +a M_x = 0; F_GF(6) + 1.875(16) - 2.00(15) = 0 F_GF = 0 + ca F_y = 0; F_GKa3 5b - 1.875 = 0 FGK = 3.125 k (C) + ca F_y = 0; 1.875 - A_y = 0 A_y = 1.875 k +a M_A = 0; B_y(32) - 4(15) = 0 B_y = 1.875 k Ax = B_x = 4 2 = 2.00 k

7–30. Solve Prob. 7–29 if the supports at A and B are pin connected instead of fixed.

D C E H I K F L J G 8 ft 8 ft 8 ft 8 ft 12 ft 3 ft 6 ft A 4 k B

(45)

2 4 4

Assume that the horizontal force components at pin supports A and B are equal. Thus,

a Ans.

a

Ans.

a Ans.

Also, referring to Fig. c, a a FDE = 9.50 k (C) :+ _{a F}_x = 0; 4 + 10.68a 8 273b - 3.125a 4 5b - 2.00 - FDE = 0 FCE = 10.68 k (T) +a M_D= 0; F_CEa 3 273b(8) - 2.00(15) = 0 FDF= 3.125 k (C) +a M_E = 0; F_DFa3 5b(8) + 1.875(8) - 2.00(15) = 0 +a M_D= 0; F_FHa3 5b(16) - 2.00(15) = 0 FFH = 3.125 k (C) FEH = 0.500 k (T) +a M_F = 0; 4(6) + 1.875(8) - 2.00(21) + F_EH(6) = 0 +a M_H = 0; F_FGa3 5b(16) + 1.875(16) - 2.00(15) = 0 FFG = 0 +a M_B = 0; A_y(32) - 4(15) = 0 A_y = 1.875 k Ax = B_x = 4 2 = 2.00 k

7–31. Draw (approximately) the moment diagram for column ACD of the portal. Assume all truss members and the columns to be pin connected at their ends. Also determine the force in members FG, FH, and EH.

D C E H I K F L J G 8 ft 8 ft 8 ft 8 ft 12 ft 3 ft 6 ft 6 ft A 4 k B

(46)

*7–32. Solve Prob. 7–31 if the supports at A and B are fixed instead of pinned.

7–31. Continued

Assume that the horizontal force components at fixed supports A and B are equal. Thus,

Also, the points of inflection N and O are 6 ft above A and B respectively. Referring to Fig. a, a Referring to Fig. b, a 1.125 - A A +a M_A= 0; M_A- 2.00(6) = 0 M_A = 12.0 k ft :+ a Fx = 0; N_x - 2.00 = 0 N_x = 2.00 k +a M_O = 0; N_y(32) - 4(9) = 0 N_y = 1.125 k Ax = B_x = 4 2 = 2.00 k D C E H I K F L J G 8 ft 8 ft 8 ft 8 ft 12 ft 3 ft 6 ft 6 ft A 4 k B

(47)

2 4 6

a Ans.

Also, referring to Fig e, a a :+ _{a F}_x = 0; 4 + 6.408 a 8 273b - 1.875a 4 5b - 2.00 - FDE = 0 FDE = 6.50 k (C) +a M_D = 0; F_CEa 3 273b(8) - 2.00(9) = 0 FCE = 6.408 k (T) +a M_E = 0; F_DFa3 5b(8) + 1.125(8) - 2.00(9) = 0 FDF= 1.875 k (C) +a M_D = 0; F_FHa3 5b(16) - 2.00(9) = 0 FFH = 1.875 k (C) +a M_F = 0; -F_EH(6) + 4(6) + 1.125(8) - 2.00(15) = 0 F_EH = 0.500 k (C) +a M_H = 0; F_FGa3 5b(16) + 1.125(16) - 2.00(9) = 0 FFG = 0 7–32. Continued

(48)

Assume the horizontal force components at pin supports A and B to be equal. Thus,

Using the method of sections, Fig. b, a Ans. a Ans. FKL = 5.286 kN (T) = 5.29 kN (T) +a M_H = 0; F_KL(1.167) + 2(0.167) + 4(1.167) + 2.889(1.5) - 3.00(5.167) = 0 FHG = 4.025 kN (C) = 4.02 kN (C) +a M_L = 0; F_HG cos 6.340° (1.167) + F_HG sin 6.340° (1.5) + 2.889(3) - 2(1) - 3.00(4) = 0 +a M_B = 0; A_y(9) - 4(4) - 2(5) = 0 A_y = 2.889 kN Ax = B_x = 2 + 4 2 = 3.00 kN

7–33. Draw (approximately) the moment diagram for column AJI of the portal. Assume all truss members and the columns to be pin connected at their ends. Also determine the force in members HG, HL, and KL.

7–32. Continued 4 kN 2 kN O K L M N A B C D E F G J I H 1.5 m 4 m 1 m 6 @ 1.5 m _{⫽ 9 m}

(49)

2 4 8

Also, referring to Fig. c, a a FJH = 7.239 kN (T) + ca F_y = 0; F_JH sin 37.87° - 14.09 sin 6.340° - 2.889 = 0 FIH = 14.09 kN (C) +a M_J = 0; F_IH cos 6.340° (1) - 2(1) - 3.00(4) = 0 FJK = 5.286 kN (T) +a M_H = 0; F_JK(1.167) + 2(0.167) + 4(1.167) + 2.889(1.5) - 3.00(5.167) = 0 7–33. Continued

(50)

Assume that the horizontal force components at fixed supports A and B are equal. Therefore,

Also, the reflection points P and R are located 2 m above A and B respectively. Referring to Fig. a

a

Referring to Fig. b,

a

Using the method of sections, Fig. d, a Ans. a Ans. Ans. FHL = 2.986 kN (C) = 2.99 kN (C) + ca F_y = 0; F_HL cos 52.13° - 2.515 sin 6.340° - 1.556 = 0 FKL = 1.857 kN (T) = 1.86 kN (T) +a M_H = 0; F_KL(1.167) + 2(0.167) + 4(1.167) + 1.556(1.5) - 3.00(3.167) = 0 FHG = 2.515 kN (C) = 2.52 kN (C) +a M_L = 0; F_HG cos 6.340° (1.167) + F_HG sin 6.340° (1.5) + 1.556(3) - 3.00(2) - 2(1) = 0 + ca F_y = 0; 1.556 - A_y = 0 A_y = 1.556 kN +a M_A = 0; M_A - 3.00(2) = 0 M_A = 6.00 kN

#

m :+ _{a F}_x = 0; P_x - 3.00 = 0 P_x = 3.00 kN +a M_R = 0; P_y(9) - 4(2) - 2(3) = 0 P_y = 1.556 kN Ax = B_x = 2 + 4 2 = 3.00 kN

7–34. Solve Prob. 7–33 if the supports at A and B are fixed instead of pinned. 4 kN 2 kN O K L M N A B C D E F G J I H 1.5 m 4 m 1 m 6 @ 1.5 m ⫽ 9 m

(51)

2 5 0

Also referring to Fig. e, a a FJH = 3.982 kN (T) + ca F_y = 0; F_JH sin 37.87° - 8.049 sin 6.340° - 1.556 = 0 FIH = 8.049 kN (C) +a M_J = 0; F_IH cos 6.340° (1) - 2(1) - 3.00(2) = 0 FJK = 1.857 kN (T) +a M_H = 0; F_JK(1.167) + 4(1.167) + 2(0.167) + 1.556(1.5) - 3.00(3.167) = 0 7–34. Continued

(52)

7–35. Use the portal method of analysis and draw the moment diagram for girder FED.

6 m A B C E D F 8 m 8 m 15 kN

(53)

2 5 2

*7–36. Use the portal method of analysis and draw the moment diagram for girder JIHGF.

15 ft A B C D E J I H G F 18 ft 18 ft 18 ft 18 ft 4 k

(54)

7–37. Use the portal method and determine (approximately)

the reactions at supports A, B, C, and D. 9 kN

5 m 5 m 5 m 4 m 4 m J K L G F E H D C B A I 12 kN

(55)

2 5 4

(56)

(57)

2 5 6

7–38. Use the cantilever method and determine (approximately) the reactions at supports A, B, C, and D. All columns have the same cross-sectional area.

9 kN 5 m 5 m 5 m 4 m 4 m J K L G F E H D C B A I 12 kN

(58)

(59)

2 5 8

7–38. Continued

7–39. Use the portal method of analysis and draw the moment diagram for column AFE.

12 ft 15 ft 4 k 5 k A D E F C B 12 ft

(60)

(61)

2 6 0

*7–40. Solve Prob. 7–39 using the cantilever method of analysis. All the columns have the same cross-sectional area. 12 ft 15 ft 4 k 5 k A D E F C B 12 ft

(62)

7–41. Use the portal method and determine (approximately) the reactions at A. 3 k 18 ft 20 ft 15 ft 15 ft G F E H D C B A I 4 k

(63)

2 6 2

7–42. Use the cantilever method and determine (approximately) the reactions at A. All of the columns have the same cross-sectional area.

3 k 18 ft 20 ft 15 ft 15 ft G F E H D C B A I 4 k

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Top story Second story Bottom story V = 3.0 k 6 + 9 + 9 - 8 V = 0; :+ _{a F}_x = 0; V = 1.875 k 6 + 9 - 8 V = 0; :+ a Fx = 0; V = 0.75 k 6 - 8 V = 0; :+ a Fx = 0;

7–43. Draw (approximately) the moment diagram for girder PQRST and column BGLQ of the building frame. Use the portal method.

6 k 9 k 9 k 15 ft 15 ft 20 ft 20 ft 10 ft 10 ft 10 ft P Q R S T K L M N O F G H I J A B C D E

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2 6 4

a a a F = 3.536 k -6(25) 9(15) 9(5) -18 33F(15) -3 33F(30) + 17 33F(50) + 37 33F(70) = 0 + a M_W = 0; F = 1.446 k -6(15) - 9(5) - 18 33F(15) -3 33F(30) + 17 33F(50) + 37 33F(70) = 0 + a M_V = 0; F = 0.3214 k - 6(5) - 18 33F(15) -3 33F(30) + 17 33F(50) + 37 33F(70) = 0 + a M_U = 0; x = 15 + 30 + 50 + 70 5 = 33 ft

*7–44. Draw (approximately) the moment diagram for girder PQRST and column BGLQ of the building frame. All columns have the same cross-sectional area. Use the cantilever method. 6 k 9 k 9 k 15 ft 15 ft 20 ft 20 ft 10 ft 10 ft 10 ft P Q R S T K L M N O F G H I J A B C D E

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The equilibrium of each segment is shown on the FBDs. V = 1.667 kN 10 - 6V = 0;

+

:a Fx = 0;

7–45. Draw the moment diagram for girder IJKL of the

building frame. Use the portal method of analysis. 20 kN

24 (10⫺3) m2 Area 16 (10⫺3) m2 _{16 (10}⫺3_{) m}2 _{24 (10}⫺3_{) m}2 4 m 5 m 4 m 4 m 4 m J K L G F E H D C B A I 40 kN

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2 6 6

The centroid of column area is in center of framework. Since , then

;

a

The equilibrium of each segment is shown on the FBDs. F1 = 1.400 k F2 = 0.359 k -2(10) - 4(F₂) + 9(F₂) + 13(3.90 F₂) = 0 +a M_M= 0; F2 = F₃ s₂= s₃; F4 = F₁ s₄= s₁; F1= 3.90 F₂ F1 12 = 6.5 2.5a F2 8 b s1= a 6.5 2.5b s2 ; s = F A

*7–46. Solve Prob. 7–45 using the cantilever method of

analysis. Each column has the cross-sectional area indicated. 20 kN

24 (10⫺3) m2 Area 16 (10⫺3) m2 16 (10⫺3) m2 24 (10⫺3) m2 4 m 5 m 4 m 4 m 4 m J K L G F E H D C B A I 40 kN